, 2020

Solution:

Suppose Sean instead follows this equivalent procedure: he starts with $M = 10\ldots 0$ on the board, as before. Instead of erasing digits, he starts writing a new number on the board. He goes through the digits of $M$ one by one from left to right, and independently copies the $n$th digit from the left with probability $\frac{1}{n}$. Now, let $a_{n}$ be the expected value of Sean's new number after he has gone through the first $n$ digits of $M$. Note that the answer to this problem will be the expected value of $a_{2021}$, since $M$ has $2021$ digits.

Note that $a_{1} = 1$, since the probability that Sean copies the first digit is $1$.

For $n > 1$, note that $a_{n}$ is $3 a_{n-1}$ with probability $\frac{1}{n}$, and is $a_{n-1}$ with probability $\frac{n-1}{n}$. Thus,
$$\mathbb{E}[a_{n}] = \frac{1}{n} \mathbb{E}[3 a_{n-1}] + \frac{n-1}{n} \mathbb{E}[a_{n-1}] = \frac{n+2}{n} \mathbb{E}[a_{n-1}].$$
Therefore,
$$\mathbb{E}[a_{2021}] = \frac{4}{2} \cdot \frac{5}{3} \cdots \frac{2023}{2021} = \frac{2022 \cdot 2023}{2 \cdot 3} = 337 \cdot 2023 = 681751$$